Answer:
it's your answer and make me barinly least
Hey can someone please help me in this question?
1 answer:
Answer:
£48
Step-by-step explanation:
Area of walls = area of four rectangular walls + area of two triangular shape at front and back
Area of a rectangular wall = l *w = 8 * 4 = 32 m²
Area of 4 rectangular walls = 4 *32 = 128 m²
Triangle:
Base = 6m
height = h = 7 - 4 = 3 m
Area of triangle = (1/2)*b*h
Area of 2 triangles = 2*9 = 18 m²
Area of four walls = 128 + 18 = 146 m²
Number of paint tins required = 146 ÷ 25 = 5.84 = 6
6 tins are required.
Cost of 6 tins = 8 * 6 = £48
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Answer:
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?
90th percentile
The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds